mit8 01sc problems25 soln

7/31/2019 MIT8 01SC Problems25 Soln

1/17

Angular Momentum ProblemsChallenge Problems

Problem 1:

Toy Locomotive A toy locomotive of mass mL runs on a horizontal circular track ofradius R and total mass m

T . The track forms the rim of an otherwise massless wheelwhich is free to rotate without friction about a vertical axis. The locomotive is startedfrom rest and accelerated without slipping to a final speed of v relative to the track. Whatis the locomotives final speed, vf , relative to the floor?

Problem 1 Solution:

We begin by choosing our system to consist of the locomotive and the track. Because

there are no external torques about the central axis, the angular momentum of the systemconsisting of the engine and the track is constant about that axis. The initial angular

momentum of the system is zero because the locomotive and the track are at rest.

Let !T

,f denote the final angular speed of the track. In the figure above the locomotive is

rotating counterclockwise as seen from above so the track must spin in the clockwise

direction. If we choose the positive z-direction to point upward then the final angularmomentum of the track is given by

! 2 L ="I ! k="m R! k (1.1),f , T,f T ,T T z T f

where we have assumed that the moment of inertia of the track about an axis passing

perpendicularly through the center of the circle forms by the track is IT z =m R2 ., T

The locomotive is moving tangentially with respect to the ground so we can choose polarcoordinates and then the final velocity of the locomotive with respect to the ground is

!v

, =vf!. (1.2)L fA point on the rim of the track has final velocity


2/17

!vT,f =#R!T,f". (1.3)

!Therefore the relative velocity v

rel =v! of the locomotive to the track is given by thedifference of the velocity of the locomotive and the appoint on the rim of the track,

! ! ! vrel =vL f v =v! ( R ! =( R != , (1.4), # T,f f # # "T,f ) vf + "T,f ) v!therefore

v =vf+R!

T,f. (1.5)

We can solve the above equation for the final angular speed !T,f in terms of the givenrelative speed and the final speed of the locomotive with respect to the ground yielding

v"vf

= . (1.6)!T,f R

The final angular momentum of the locomotive with respect to the center of the circle

formed by the track when it is moving with speed vf , relative to the floor is

!L =m Rv k

, L . (1.7)L f fBecause the angular momentum of the system is constant we have that

! ! ! 2 0=L +L =("m R! +mRv )k. (1.8)T f, L f, T T,f L fNow substitute Eq. (1.6) into the z-component of the above equation yielding

0 =!mTR(v !v

f) +m

LRv

f. (1.9)

We now solve the above equation to find the final speed of the locomotive relative to thefloor

mT v . (1.10)

mT+m

L

vf =


3/17

Problem 2: Spinning Chair and Wheel Solution

A person is sitting on a stool that is initially not rotating and is holding a spinning wheel.The moment of inertia of the person and the stool about a vertical axis passing through

the center of the stool is IS,p

. The moment of inertia of the wheel about an axis,

perpendicular to the plane of the wheel, passing through the center of mass of the wheelis I =(1 / 4)I

S,p. The mass of wheel is m . Suppose that the person holds the wheel as

w w

shown in the sketch such that the distance of an axis passing through the center of mass

of the wheel to the axis of rotation of the stool is d and that mwd

2=(1/ 3)I

w. Suppose

the wheel is spinning initially at an angular speed !s

. The person then turns the spinning

wheel upside down. You may ignore any frictional torque in the bearings of the stool.

What is the angular speed of the person and stool after the spinning wheel is turned

upside down?

Problem 2 Solution:

There are no external torques acting on the system so angular momentum is conserved.

When the person is spinning on the chair, the angular momentum about a vertical axis (z-axis) passing through the center of the stool is the sum of two contributions. The first

contribution is due to the angular momentum of the person and the stool,

!L =I !kS p, ,S p


4/17

!

where =!k is the angular velocity of the person and the stool about a vertical axispassing through the center of the stool.

The second contribution to the angular momentum about a vertical axis passing through

the center of the stool is due to the rotational motion of the wheel. From the parallel axis

theorem, the moment of inertia of the wheel about a vertical axis passing through thecenter of the stool is

d2

I =I +m .S,w w w

!

Since the wheel is rotating about this axis with angular velocity =!k , the angularmomentum is given by

!rot 2 L

, = Iw+m d !k.S w ( w )Note the first term represent s the fact that when the wheel rotates once about the axis, it

is also rotating once about its center of mass. The second term can be thought of as theangular momentum of all the mass of the wheel located at its center of mass undergoing

uniform circular motion.

Since the wheel is also spinning about its center of mass with angular velocity!

spin =!sk, there is the spin angular momentum due entirely to the spinning of thewheel about its center of mass,

!

spin L =I !k.w w s

So the total angular momentum of the wheel about the central axis is

! ! !total rot spin 2 L , S w+L , I +m d )! I !S w =L , cm w = (( w w + w s)k

The total angular momentum of the person, stool, and wheel, is then the sum of thesethree terms.

! ! ! !total rot spin 2 L =LS p +L +L = (I , +I +m d )!+I ! kS , S w, w ( s p w w w s)

Initially, !i=0 , therefore the initial angular momentum is only due to the spin angular

momentum of the wheel,

!

totalL

S i, =Iw!sk


5/17

When the person turns the wheel over, the spin angular momentum of the wheel has

reversed direction,

!

spin L ="I !k.w f, w sThe system begins to rotate with angular velocity!

f, so the final angular momentum is

!

total 2 L, =(I , +Iw+m d )!f k"I !sk.S f s p w w

Notice that the sign of the spin angular momentum of the wheel has changed. Sinceangular momentum is conserved,

! !

total totalL =L ,S i, S f,

we have that (for the z-component of angular momentum)

I! =(I +I +m d2 )! "I !w s s,p w w f w s

We can solve for!f

:

2I! 2I! 3w s w s

!f= = = !

s(I +I +m d2 ) (I +I +m d2 ) 8

s,p w w s,p w w


6/17

Problem 3:

A drum of mass mA and radius a rotates freely with initial angular velocity !A,0. A

second drum with mass mB and radius b>a is mounted on the same axle and is at rest,

although it is free to rotate. A thin layer of sand with mass mS is distributed on the inner

surface of the smaller drum. At t=0, small perforations in the inner drum are opened.The sand starts to fly out at a constant rate !kg"s-1 and sticks to the outer drum. Find thesubsequent angular velocities of the two drums !

A and !B. Ignore the transit time of thesand.

Problem 3 Solution:

In this problem, we choose as our system the outside drum with accumulated sand andthe small amount of sand m

S that leaves the inner drum and settles against the outerdrum, with coordinate system shown below.

The angular impulse equation states that

!

!#!Sdt="LS . (3.1)


7/17

As the sand leaves the inner drum it does not exert any force on the inner drum, it fliesthrough the open whole, hence the inner drum does not exert any force or torque about

the center S of the drums on the system consisting of the sand and outer drum. Thereforethe angular momentum about the center Sof the drums is constant.

! !LS i, =LS f, . (3.2)

The initial angular momentum about S is

!2

S i m a k. (3.3)L , = s !A,0When all the sand has reached the outer drum, the final angular momentum about S is

!LS f, =(ms+mB)b2!B f, k. (3.4)

The z-component of the angular momentum equation (Eq. (3.2)) becomes

m a2! =(m +m )b2! . (3.5)s A,0 s B B,f

Therefore the final angular speed of the outer drum is given by

m a! = s 2!A,0 . (3.6)

B,f(m +m

B)b2

s

The inner drum does not change its angular speed.

Note that at time t, an amount of sand ms(t) =!t has been transferred so the

angular momentum at time t is given by

!2 2 L

st a

B ) ( )S( )t =(m #! ) "A,0k+(m +!t b"B t k. (3.7)So the angular momentum equation becomes

msa

2! =(m

s" #t)a2! +(m

B+#t)b2! (t) . (3.8)

A,0 A,0 B

which we can solve for the angular speed of the outer drum as a function of time

"ta2!

A,0!

B(t) = . (3.9)

("t+m )b2B


8/17

Problem 4: Measuring Moment of Inertia

In the angular momentum experiment, shown tothe right, a washer is dropped smooth side down

onto the spinning rotor.

The graph below shows the rotor angular velocity"

!(rad s# 1)as a function of time.Assume the following:

The rotor and washer have the samemoment of inertia I.

!

The friction torque !f on the rotor isconstant during the measurement.

!

a) Find an expression for the magnitude !f in terms of I and numbers you obtainfrom the graph.

b) What torque does the rotor exert on the washer during the collision (betweent=1.90s and t=2.40s)? Express your answer in terms of I and numbers youobtain from the graph.

c) How many radians does the rotor rotate during the collision (betweent=1.90s and t=2.40s)? Give a numerical answer.

d) How many radians does the washer rotate during the collision (betweent=1.90s and t=2.40s)? Give a numerical answer.Note: express all of your answers in terms of I and numbers you obtain from the graph.Be sure to give an analytic expression prior to substituting the numbers from the graph.


9/17

Problem 4 Solutions:

a) First, make sure that the problem makes sense. Between times t=0.40s andt=1.90s, the magnitude of the angular acceleration is #!/#t=40rad s$ "2 and betweentimes t=2.40s and t=4.40s the magnitude of the angular acceleration is#!/#t=20rad s"2. During these two time intervals, the only torque is the friction$torque, assumed constant, and doubling the net moment of inertia halves the angular

acceleration.!

" ! "We then have ! =I(40rad s# 2). This is also ! =2I(20rad s# 2), but thats not partf fof this problem, just a consistency check.

b) The main point to recognize in this part of the problem is that the washer acceleratesfrom an initial angular velocity of zero to the final angular velocity shown on the graph.

The only torque on the washer is the torque that the rotor exerts on the washer, and themagnitude of this torque is

$! 100rad s% #1I(200rad s#2) (4.1)washer" =I =I = %rotor-washer

$t 0.50s

c) Lets go the simple way, and say that the angle through which the rotor rotates is the

product of the average angular velocity and the time interval,

$!=" $t= (160rad s )(0.50s)=80rad,ave % #1 (4.2)about 50revolutions.d) Weve done part c) simply, so lets do the same for the washer. Here, the average

angular velocity is 50rad s!1 over the same time interval, or 25rad."


10/17

Problem 5

In the angular momentum experiment, shown tothe right, a washer is dropped smooth side down

onto the spinning rotor.

The graph below shows the rotor angular velocity"

!(rad s# 1)as a function of time.Assume the following:

The rotor and washer have the samemoment of inertia I.

!

The friction torque !f on the rotor isconstant during the measurement.

Note: express all of your answers in terms of I and numbers you obtain from the graph.Be sure to give an analytic expression prior to substituting the numbers from the graph.

!

a. Find an expression for the magnitude !f in terms of I and numbers you obtainfrom the graph.

b. How much mechanical energy is lost to bearing friction during the collision(between t=1.90s and t=2.40s)?

c. How much mechanical energy is lost to friction between the rotor and the washerduring the collision (between t=1.90s and t=2.40s)?


11/17


a) First, make sure that the problem makes sense. Between times t=0.40s andt=1.90s, the magnitude of the angular acceleration is #!/#t=40rad s$ "2 and betweentimes t=2.40s and t= 4.40s the magnitude of the angular acceleration is#!/#t=20rad s"2. During these two time intervals, the only torque is the friction$torque, assumed constant, and doubling the net moment of inertia halves the angular

acceleration.!

" ! "We then have !f =I(40rad s# 2). This is also !f =2I(20rad s# 2), but thats not partof this problem, just a consistency check.

"1 "1For parts (b) and (c), denote ! =220rad s# , ! =100rad s# , so thatinitial final

1 2 2 "2K = I!

initial =I(24,200rad #s )initial21 2 2 "2

K = (2I)! =I(10,000rad #s ).final final2

b) The mechanical energy lost due to the bearing friction is the product of the magnitude

of the frictional torque and the total angle "! through which the bearing has turnedduring the collision. A quick way to calculate "! is to use

#1$!=" $t= (160rad s )(0.50s)=80rad,ave %

! 2)so #$E =! $"=I(3200rad .bearing f

c) The energy lost due to friction between the rotor and the washer is then

2 2!"K+!"Ebearing =Kinitial!Kfinal!I(3200rad )=I(11,000rad ).


12/17

Problem 6: Work Done by Frictional Torque

A steel washer is mounted on the shaft of a small motor. The moment of inertia of the

motor and washer is I0

. The washer is set into motion. When it reaches an initial angular

velocity !0, at t=0 , the power to the motor is shut off, and the washer slows down

during a time interval !t =t until it reaches an angular velocity of ! at time t . At1 a a a

that instant, a second steel washer with a moment of inertia Iw

is dropped on top of the

first washer. Assume that the second washer is only in contact with the first washer. The

collision takes place over a time !tcol=t

b"t

aafter which the two washers and rotor

rotate with angular speed !b. Assume the frictional torque on the axle is independent of

speed, and remains the same when the second washer is dropped.

a) What angle does the rotor rotate through during the collision?

b) What is the work done by the friction torque !f from the bearings during thecollision?


During the collision, the component of the average angular acceleration of the rotor isgiven by

"b#"a


13/17

So the work done by the bearing friction during the collision is

1 $!0 "!a'

= I0Wnc,b ()(!a+!b)#tint


14/17

Problem 7: Experiment Angular Momentum:Angular Collision

A steel washer is mounted on the shaft of a small motor. The moment of inertia of the

motor and washer is I0

. The washer is set into motion. When it reaches an initial angular

velocity !0, at t=0 , the power to the motor is shut off, and the washer slows down until

it reaches an angular velocity of!a

at time ta. At that instant, a second steel washer with

a moment of inertia Iw

is dropped on top of the first washer. Assume that the second

washer is only in contact with the first washer. The collision takes place over a time

=tb"t . Assume the frictional torque on the axle is independent of speed, and!t

col a

remains the same when the second washer is dropped. The two washers continue to slow

down during the time interval !t2=t

f"t

buntil they stop at time t=t

f. Your answers

should be in terms of all or any of I0

, Iw

, !0, !

a, and !t

2(these would be, !t

1, !t

col

measured or observed values).

a) What is the angular deceleration !1

while the washer and motor are slowing

down during the interval !t1=t

a?

b) What is the angular impulse during the collision?c) What is the angular velocity !

bof the two washers immediately after the

collision is finished (when the washers rotate together)?d) What is the angular deceleration !

2after the collision?


a) The angular acceleration of the motor and washer from the instant when the power isshut off until the second washer was dropped is given by

"a#"

0 . (7.1)!1=

$t1

b) The angular acceleration found in part a) is due to the frictional torque in the motor,

I0(#

a$#

0)

. (7.2)=I0"

1=!

friction%t

1


15/17

Note that because !0>! therefore !

1is negative, as indicated

a


16/17

d) The final angular acceleration !2

is given by

%% ( (#b

1 I0 $tcol!

2=" =" *(#a "#0 ) +#a *.''

$t2 $

t2 (

I0+I

) $t1 ) )w &&


17/17

MIT OpenCourseWarehttp://ocw.mit.edu

8.01SC Physics I: Classical Mechanics

For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms.

mit8 01sc problems25 soln

Documents